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2003 AIME II problem 10

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Two positive integers differ by 60. The sum of their square roots is the square root of an integer that is not a perfect square. What is the maximum possible sum of the two integers? So the second sentence is a little confusing, but let's take a step-by-step. So two positive integers differ by 60. So let's say they're a and b. So we have a and b. They're both greater than 0. They're also both integers. And they differ by 60. So a minus b is equal to 60. Or we could add b to both sides, and we could get a is equal to b plus 60. They're 60 apart. Fair enough. Now this second sentence, the sum of their square roots-- the square root of a plus square root of b-- the sum of the square roots is the square root of an integer that is not a perfect square. So it's equal to the square root of some integer. So c is an integer, but it's not a perfect square. So square root of c is not an integer. So that's the second part. What is the maximum possible sum of the two integers? So if we could maximize b and find an a that fits it, or an a and a b that meet these constraints we could get them as big as possible, then the sum of a and b will also be maximized. So let's think about that in a second. But just to simplify-- this is actually the really confusing part because c is an integer and square root of c is not an integer. And to make it a little less confusing, let's just square both sides of this equation over here. So square root of a squared is-- so we're just going to square both sides. So this is going to be square root of a times square root of a-- which is a-- plus 2 times the square root of a times the square root of b-- which is 2 times the square root of ab-- plus the square root of b squared-- which is just going to be b-- is equal to c. Now, we know that a is an integer. We know that b is an integer. And we know that c is an integer. We know that these are integers. So if I have an integer plus something plus an integer equaling another integer, this thing in the middle has to be an integer. And actually, I'll show you right now that not only does this thing have to be an integer, square root of ab has to be an integer. And you might be saying, wait, wait, wait, wait a second. If this thing just has to be an integer, can't I just say-- let me do it over here on the side-- can't I just say that 2 times the square root of ab has to be equal to some integer. You could divide both sides by 2. And you could say, well, maybe the square root of ab is equal to k/2, where this is not an integer. I claim that this thing does have to be an integer. But you say, hey look, just straight from the math, this could be an integer. But maybe divide both sides by 2. And this doesn't have to be an integer. And if this is not an integer, that means that k is odd, because otherwise it would be divisible by 2. Now, if we square both sides of this, we would get ab is equal to k squared over 4. Well, this is going to be an odd times an odd. It's going to be an odd squared. So this is still going to be odd, which is not divisible by 4. So this thing on the right hand side will still not be an integer. But remember, a and b are integers. So a times b, this is an integer. So if a and b are integers, there's no way that the square root of its product can be something divided by 2, where this doesn't reduce any more, where this doesn't simplify, or some odd number divided by 2. So we just get a contradiction here. Or another way to say it is the square root of ab-- because this whole thing has to be an integer and both a and b are integers, the square root of ab has to be an integer. But if the square root of ab has to be an integer, then that means that ab has to be a perfect square. So then I'll start over here so I don't run out of space. Actually, let me do it right under that because it's a straight result. So that means that a times b is a perfect square. So let's see how we can deal with this a little bit. We can actually do a substitution for a. a is equal to b plus 60. So this means that b plus 60 times b is equal to a perfect square, so it could be equal to some number n squared, where n is an integer. And let's see if we can play around with this a little bit, put some constraints on what we're doing. So if you multiply this out, you get b squared plus 60b-- I'll leave some space over here-- is equal to n squared. And let's see, we could complete the square over here. And maybe we could subtract this over and we'll have a difference of squares. So we want to complete the square here. You take half of this. So half of 60, half of the coefficient-- that's 30-- and square it. So this would be plus 900 minus 900. We haven't changed it. We're adding and subtracting the same thing. But what that does for us is that this part over here becomes b plus 30 squared, and then we have the minus 900 is equal to n squared. And then we could subtract n squared from both sides and add 900 to both sides. And we get b plus 30-- and you really just have to play around with this thing and see what constraints you can get. b plus 30 squared minus n squared is equal to 900. And this is just a difference of squares. So this can be factored. This is b plus 30. This is a plus b. Or, I shouldn't use a and b. But it's the square root of this plus the square root of this, so plus n. Times the square root of this minus the square root of this. So, b plus 30 minus n. So we can factor it just like that. And that's about as good of a constraint as we can get. Remember, we want to be able to maximize b. If we're maximizing b, a is only 60 above b. So we're also going to be maximizing a. And so we'll also be maximizing the sum. So we want b to be as high as possible. So b plus 30 plus n is going to be some factor of 900. And b plus 30 minus n will be the other factor of 900. And we could actually do a little bit more of an insight here. We have two factors. Their product is an even number. So it's an even product. And if we were to take the sum of these two factors, if you were to take b plus 30 plus n and add it to b plus 30 minus n-- so we're just taking the sum of these two factors-- we get 2b plus 60. And these cancel out. And so this is also even. So if I take two numbers, and their product is even, and their sum is also even, that means the two numbers have to be even. So both this guy and this guy over here have to be even. Both of them have to be even. And at this point, I think we've used up all the information in the problem. We really just have to try out factors of 900 that meet these constraints where both are even. And let's try with the most extreme things in an attempt to maximize our b. So let's think about the factors of 900. You could do 1 and 900, but remember, both factors have to be even. 1 is not even. We could try 2 and 450, where this will be the higher factor. This will be the lower factor because we're subtracting a number here. So let's try that out. Actually, let me do it over here. Actually, I'll do it over here. Maybe this is 450 and this is 2. So we have b plus 30 plus n is equal to 450, and that b plus 30 minus n is equal to 2. And now we can solve for b. We can add these two equations. You get 2b plus 60-- these cancel out-- is equal to 452. Subtract 60 from both sides. 2b is equal to 392, or b is equal to 196. So let's see how this works out. If b is equal to 196, then a is just 60 plus that. So a is equal to 256. And let's see what it does to this equation over here. Square root of a is 16. So we would have the square root of 256 plus the square root of 196. Well, that's equal to 16 plus 14, which is equal to 30. So it looks like it might work. But remember, this thing over here, the square root of c, is not a perfect square. This value right over here. That's the square root of c. Or I should say the square root of c is not an integer. It cannot be an integer. If it was, then c which in this case would be 900, would be a perfect square. But they told us in the problem c is not a perfect square. So this one actually does not work because this would give us a c that is a perfect square. It gives us 900. Square root of c is 30. So we have to cross these out too, although it was a very tantalizing solution since it gave us nice clean numbers. So let's keep trying factors of 900. So remember, you could try 3 and 300, but we know that they both have to be even. So we could just reject this because 3 is not even. So maybe it's 4 and-- 4 goes into 900 125 times. You might be tempted to try these out. But 125 is not even. So we could keep going. So then we could go to 5. Well, once again 5 is not even so that won't work. We could try 6. 6 goes into 900-- it goes into 90 15 times. So it goes into 900 150 times. They're both even numbers. So maybe, just maybe these will work. So let's try it out again. So let me scroll down a little bit. So you could have b plus 30 plus n-- that'll be the larger number-- is equal to 150 because we're adding the n there. And then you have b plus 30 minus n would be equal to 6. You add both equations, you get 2b plus 60 is equal to 156. Or subtracting 60 from both sides 2b is equal to-- this is equal to 96. Or b is equal to 48. So here's our next candidate. Our next candidate is b is equal to 48. Then a would be 60 more than this. a would be equal to 108, 60 more than that. If I were to take the square root of 108 plus the square root of 48, what does that give me? 108 is 36 times 3. So this is the square root of 36 times the square root of 3, or 6 square roots of 3. Plus-- this is 16 times 3. So it's going to be the square root of-- well 48 is 16 times 3. So it's the square root of 16 times the square root of 3. So it's 4 square roots of 3. So when you add these two together, you get 10 square roots of 3. So this fits our bill for the square root of c. So over here we have the square root of c is equal to 10 roots of 3. It's definitely not an integer. So it makes that. And if you square it, you would get c as being equal to 100 times 3, which is equal to 300. And it's an integer. So it works. So our best shot, our best values for a and b, b is 48, a is 108. If we take the sum, 108 plus 48, we get 16, 5, and 156. And you can see when you keep trying-- obviously we didn't have the best result when we had b being 196. But as we get our factors closer and closer together, the b is going to go lower and lower and lower. And we're obviously going to have a less of a maximum value for a plus b. So as we got from the most extreme factors of 900, we keep going until something met. This met our constraints. If we kept going, if we tried eight and-- I don't know if that would even work as a factor. But if we just tried and kept looking at the other factors, we actually would have gotten numbers that are closer to each other. And that would have led to lower values for our b. So this right here is our maximum b-- or the b that will maximize our result. And the sum is 156.